# Exercise 8.1.3

Show that there is no comparison sort whose running time is linear for at least half of the $n!$ inputs of length $n$. What about a fraction $1/n$ of the inputs of length $n$? What about a fraction $1/2^n$?

If it is linear for at least half of the inputs, then using the same reasoning as in the text, it must hold that:

$$ \frac{n!}{2} \le 2^n $$

This holds only for small values for $n$. Same goes for the other:

$$ \frac{n!}{n} \le 2^n $$

And:

$$ \frac{n!}{2^n} \le 2^n \Leftrightarrow n! \le 4^n $$

All those have solutions for small $n < n_0$, but don't hold for larger values.

In contrast, insertion sort gets its work done in $\Theta(n)$ time in the best case. But this is a $1/n!$ fraction of the inputs, which is smaller than $1/2^n$.